Test Index

TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Section: Mathematics

© examsnet.com

Question : 77 of 160

Marks: +1, -0

For

-1 < x < 1

y>1

\int \frac{x}{\sqrt{1+x}+\sqrt{1-x}}

dx

\int \frac{y}{\sqrt{y+1}+\sqrt{y-1}}

dy

A(1+x)^{3/2}+B(1-x)^{3/2}+f(y)(y+1)^{3/2}+g(y)(y-1)^{3/2}+C

Af(y)+Bg(y)

=

$\frac{2y}{15}$
$-\frac{4}{45}$
$-\frac{4}{15}$
$\frac{3y+2}{45}$

Solution: 👈: Video Solution

\int\;\frac{x}{\sqrt{1+x}+\sqrt{1-x}} dx+\int\;\frac{y}{\sqrt{y+1}+\sqrt{y-1}} dy

\int\;\frac{x(\sqrt{1+x}-\sqrt{1-x})}{(1+x)-(1-x)} dx+\int\;\frac{y(\sqrt{y+1}-\sqrt{y-1})}{(y+1)-(y-1)} dy

\frac{1}{2} \int (\sqrt{1+x}-\sqrt{1-x}) dx+\frac{1}{2} \int y\sqrt{y+1}-y\sqrt{y-1} dy

\frac{(x+1)^{3/2}}{3}+\frac{(1-x)^{3/2}}{3}+\frac{1}{2} \int y\sqrt{y+1}-y\sqrt{y-1} dy

. . . (i)

For

\int y\sqrt{y+1} dy \Rightarrow y+1=t^{2}

On differentiate,

dy=2t dt

\;\int (t^{2}-1) t (2t dt)

\;\; =2 \int (t^{4}-t^{2}) dt=2\left(\frac{t^{5}}{5}-\frac{t^{3}}{3}\right)+C

For

\;\int y\sqrt{y-1} dy

y-1=t^{2} \Rightarrow dy=2t dt

\;\int (t^{2}+1) t \cdot 2t dt

\;\; =2 \int (t^{4}+t^{2}) dt=\frac{2}{5} t^{5}+\frac{2}{3} t^{3}+C

Now, from Eq. (i)

\frac{(x+1)^{3/2}}{3}+\frac{(1-x)^{3/2}}{3}+\frac{(y+1)^{5/2}}{5}-\frac{(y-1)^{3/2}}{3}

-\frac{(y-1)^{5/2}}{5}-\frac{(y-1)^{3/2}}{3}

(y+1)^{3/2} \underbrace{\left(\frac{y+1}{5}-\frac{1}{3}\right)}_{f(y)}+(y-1)^{3/2} \underbrace{\left[-\frac{y-1}{5}-\frac{1}{3}\right]}_{g(y)}

A=\frac{1}{3}

and

B=\frac{1}{3}

\frac{1}{3}\left(\frac{y+1}{5}-\frac{1}{3}\right)-\frac{1}{3}\left(\frac{y-1}{5}-\frac{1}{3}\right)=\frac{-4}{45}

© examsnet.com

Go to Question:

Prev Question

Next Question